Question Number 75465 by necxxx last updated on 11/Dec/19 Commented by necxxx last updated on 12/Dec/19 $${Find}\:{the}\:{yellow}\:{area} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 9928 by Tawakalitu ayo mi last updated on 16/Jan/17 Commented by RasheedSoomro last updated on 16/Jan/17 $$\mathrm{Picture}\:\mathrm{not}\:\mathrm{visible}. \\ $$ Terms of Service Privacy…
Question Number 140997 by loveineq last updated on 14/May/21 $$\mathrm{Let}\:{a},{b},{c}\:\geqslant\:\mathrm{0}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{8}}\centerdot\frac{\left(\mathrm{2}+{a}\right)\left(\mathrm{2}+{b}\right)\left(\mathrm{2}+{c}\right)}{\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)}\:\geqslant\:\frac{\mathrm{4}−{a}−{b}−{c}}{\mathrm{4}+{a}+{b}+{c}}\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$ Commented by MJS_new last updated on 15/May/21 $$\mathrm{as}\:\mathrm{easy}\:\mathrm{as}\:\mathrm{the}\:\mathrm{last}\:\mathrm{one} \\…
Question Number 140996 by mnjuly1970 last updated on 14/May/21 $$ \\ $$$$\:\:\:\:\:\:\mathscr{E}{valuation}\:{of}\:::\:\Omega\::=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{1}+{n}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:{solution}:: \\ $$$$\:\:\:\:\:\Omega:=\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} −{i}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\underset{{n}=\mathrm{1}}…
Question Number 75463 by indalecioneves last updated on 11/Dec/19 Answered by mr W last updated on 12/Dec/19 $${shape}\:{of}\:{cable}\:{is}\:{catenary}: \\ $$$$\frac{{L}}{{d}}=\frac{\mathrm{2}\:\mathrm{sinh}\:\frac{{d}}{{x}}}{\frac{{d}}{{x}}} \\ $$$$\frac{{f}}{{d}}=\frac{\mathrm{cosh}\:\frac{{d}}{{x}}−\mathrm{1}}{\frac{{d}}{{x}}} \\ $$$${with}\:{t}=\frac{{d}}{{x}}\:{as}\:{parameter}: \\…
Question Number 140998 by mohammad17 last updated on 14/May/21 Answered by Ar Brandon last updated on 14/May/21 $$\mathcal{I}=\int\frac{\mathrm{sin4x}}{\mathrm{cos}^{\mathrm{4}} \mathrm{x}+\mathrm{4}}\mathrm{dx}=\int\frac{\mathrm{2sin2xcos2x}}{\left(\frac{\mathrm{1}+\mathrm{cos2x}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{4}}\mathrm{dx},\:\mathrm{c}=\mathrm{cos2x} \\ $$$$\:\:\:=−\int\frac{\mathrm{c}}{\frac{\left(\mathrm{c}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}+\mathrm{4}}\mathrm{dc}=−\mathrm{4}\int\frac{\mathrm{cdc}}{\mathrm{c}^{\mathrm{2}} +\mathrm{2c}+\mathrm{17}}\mathrm{dc} \\…
Question Number 9923 by konen last updated on 16/Jan/17 $$\mathrm{i}=\sqrt{−\mathrm{1}} \\ $$$$\mathrm{z}=\:\left(\mathrm{1}−\mathrm{i}\right)^{\mathrm{17}} −\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{17}} \:\Rightarrow\:\mathrm{z}=?\: \\ $$ Commented by RasheedSoomro last updated on 16/Jan/17 $$\mathrm{z}=\:\left(\mathrm{1}−\mathrm{i}\right)^{\mathrm{17}} −\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{17}}…
Question Number 75456 by peter frank last updated on 11/Dec/19 $${Find}\:{the}\:{domain}\:{and}\: \\ $$$${range}\:{of}\:{relation} \\ $$$$\left({a}\right){R}=\left\{\left({x},{y}\right):{y}=\sqrt{{x}^{\mathrm{2}} −\mathrm{6}}\:\right\} \\ $$$$\left({b}\right){R}=\left\{\left({x},{y}\right):{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{1}\:\right\} \\ $$ Answered by Kunal12588…
Question Number 9921 by Tawakalitu ayo mi last updated on 15/Jan/17 Answered by mrW1 last updated on 16/Jan/17 $$\mathrm{2}{x}=\mathrm{180}−\mathrm{50}−\mathrm{70}=\mathrm{60} \\ $$$$\Rightarrow{x}=\mathrm{30} \\ $$$${y}=\mathrm{180}−\mathrm{50}−{x}=\mathrm{100} \\ $$$${z}=\mathrm{180}−{x}−\left(\mathrm{70}+\mathrm{50}\right)=\mathrm{180}−\mathrm{30}−\mathrm{120}=\mathrm{30}…
Question Number 75457 by Master last updated on 11/Dec/19 Commented by MJS last updated on 11/Dec/19 $$\mathrm{no}\:\mathrm{exact}\:\mathrm{solution}\:\mathrm{possible} \\ $$$$\mathrm{try}\:\mathrm{to}\:\mathrm{approximate},\:\mathrm{there}\:\mathrm{are}\:\mathrm{3}\:\mathrm{solutions}, \\ $$$$\mathrm{one}\:\mathrm{real}\:\mathrm{and}\:\mathrm{two}\:\mathrm{conjugated}\:\mathrm{complex} \\ $$ Commented by…